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A244940
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T(n,k)=Number of length n 0..k arrays with each partial sum starting from the beginning no more than sqrt(3) standard deviations from its mean
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14
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2, 3, 4, 4, 9, 8, 5, 14, 25, 14, 6, 23, 52, 69, 28, 7, 34, 111, 198, 207, 56, 8, 43, 190, 517, 764, 603, 104, 9, 58, 295, 1076, 2529, 2976, 1741, 208, 10, 75, 444, 1939, 6370, 12497, 11668, 5223, 416, 11, 94, 631, 3358, 13139, 37364, 60773, 45960, 15445, 796, 12, 109
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OFFSET
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1,1
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COMMENTS
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Table starts
...2.....3......4.......5........6.........7.........8..........9.........10
...4.....9.....14......23.......34........43........58.........75.........94
...8....25.....52.....111......190.......295.......444........631........896
..14....69....198.....517.....1076......1939......3358.......5405.......8450
..28...207....764....2529.....6370.....13139.....26256......47837......81956
..56...603...2976...12497....37364.....89937....203496.....414005.....802938
.104..1741..11668...60773...219382....619567...1606524....3633503....7812680
.208..5223..45960..293467..1290578...4214681..12610616...32062641...76978472
.416.15445.181652.1452027..7608118..29077603..98974880..284037099..754859772
.796.45423.719784.7098491.44939408.198628937.777361848.2523617923.7399299882
Computation in integer form, using 6 times the 0..k mean and 36 times the variance, mean6(k)=3*k; var36(k)=6*k*(2*k+1)-mean6(k)^2; then (6*sum{x(i),i=1..j}-j*mean6(k))^2<=3*j*var36(k) for all j=1..n
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LINKS
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EXAMPLE
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Some solutions for n=6 k=4
..4....1....2....0....4....3....3....0....1....1....3....1....0....4....3....4
..1....3....1....4....0....0....1....3....3....0....4....4....4....1....3....1
..3....1....0....2....4....4....4....2....3....3....3....0....2....2....0....1
..3....3....4....3....3....2....4....1....1....2....2....4....4....1....3....4
..4....1....1....2....3....4....0....1....1....2....3....4....3....0....0....0
..3....3....0....4....3....3....0....0....1....3....0....4....2....1....3....4
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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